HMMT 二月 2023 · 冲刺赛 · 第 21 题

HMMT February 2023 — Guts Round — Problem 21

[18] Let x , y , and N be real numbers, with y nonzero, such that the sets { ( x + y ) , ( x − y ) , xy, x/y } and { 4 , 12 . 8 , 28 . 8 , N } are equal. Compute the sum of the possible values of N .

解析

[18] Let x , y , and N be real numbers, with y nonzero, such that the sets { ( x + y ) , ( x − y ) , xy, x/y } and { 4 , 12 . 8 , 28 . 8 , N } are equal. Compute the sum of the possible values of N . Proposed by: Sean Li 426 Answer: 5 Solution: First, suppose that x and y were of different signs. Then xy < 0 and x/y < 0, but the set has at most one negative value, a contradiction. Hence, x and y have the same sign; without loss of generality, we say x and y are both positive. 2 2 1 2 2 s + d Let ( s, d ) := ( x + y, x − y ). Then the set given is equal to { s , d , ( s − d ) , } . We split into two 4 s − d cases: s + d 1 2 2 • Case 1: = N . This forces s = 28 . 8 and d = 12 . 8, since (28 . 8 − 12 . 8) = 4. Then s − d 4 √ √ 12+8 12 − 8 s = 12 0 . 2 and d = ± 8 0 . 2, so N is either = 5 or = 0 . 2. 12 − 8 12+8 s + d s + d 2 2 • Case 2: 6 = N . Suppose = k , so ( s, d ) = (( k + 1) t, ( k − 1) t ) for some t . Then s : d : s − d s − d 1 2 2 2 2 ( s − d ) = ( k + 1) : ( k − 1) : k . Trying k = 4 , 12 . 8 , 28 . 8 reveals that only k = 4 is possible, 4 2 5 2 2 since 28 . 8 : 12 . 8 = (4 − 1) : 4. This forces N = s = · 12 . 8 = 80. 4 Hence, our final total is 5 + 0 . 2 + 80 = 85 . 2.